Answer:
146.41
Step-by-step explanation:
third order determinant = determinant of 3×3 matrix A
given ∣A∣=11
det (cofactor matrix of A) =set (transpare of cofactor amtrix of A) (transpare does not change the det)
=det(adjacent of A)
{det (cofactor matrix of A)} ^2 = {det (adjacent of A)}
^2
(Using for an n×n det (cofactor matrix of A)=det (A)^n−1
)
we get
det (cofactor matrix of A)^2 = {det(A) ^3−1
}^2
=(11)^2×2 = 11^4
=146.41
Answer:
Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Step-by-step explanation:
We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.
So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;
Z =
~ N(0,1)
where,
= average age of the random sample of horses with colic = 12 yrs
= average age of all horses seen at the veterinary clinic = 10 yrs
= standard deviation of all horses coming to the veterinary clinic = 8 yrs
n = sample of horses = 60
So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(
12)
P(
12) = P(
) = P(Z
1.94) = 1 - P(Z < 1.94)
= 1 - 0.97381 = 0.0262
Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
The inequality can be:

Now you divide both sides by 150 to get the #of boxes.

Lastly, x is 20. So the maximum amount of boxes that the freight elevator can hold is 20
Answer:
DEC+DEF=180
DEC=180-116
DEC=64°
in triangle DCE
angle D+angle C+angle E=180
7y+6+4y+64=180
11y+70=180
11y=180-70
11y=110
y=110/11
y=10°
angle C=4y
=4(10)
=40°
Answer:
9/5 (K-273.15) + 32=F
Step-by-step explanation:
K=5 /9(F−32)+273.15
Subtract 273.15 from each side
K-273.15=5/9(F−32)+273.15-273.15
K-273.15=5/9(F−32)
Multiply by 9/5 on each side
9/5 (K-273.15)= 9/5 *5/9(F−32)
9/5 (K-273.15)=(F−32)
Add 32 to each side
9/5 (K-273.15) + 32=F−32 +32
9/5 (K-273.15) + 32=F